Question

A particle moving with uniform acceleration has average velocities $v_1$, $v_2$ and $v_3$ over the successive intervals of time $t_1$, $t_2$ and $t_3$ respectively. The value of $\frac{\left(v_{1}-v_{2}\right)}{\left(v_{2}-v_{3}\right)}$ will be

• A. $\frac{t_{1}-t_{2}}{t_{2}-t_{3}}$
• B. $\frac{t_{1}-t_{2}}{t_{2}+t_{3}}$
• C. $\frac{t_{1}+t_{2}}{t_{2}-t_{3}}$
• D. $\frac{t_{1}+t_{2}}{t_{2}+t_{3}}$

Answer 1

Answer: (D)

Let $u$ be initial velocity and $a$ be uniform acceleration.

Average velocities in the intervals from $0$ to $t_1$, $t_1$ to $t_2$ and $t_2$ to $t_3$ are
$v_{1}=\frac{u+u+at_{1}}{2}=u+\frac{a}{2}t_{1}\quad\ldots\left(i\right)$
$v_{2}=\frac{u+at_{1}+u+a\left(t_{1}+t_{2}\right)}{2}=u+at_{1}+\frac{a}{2}t_{2}\quad\ldots\left(ii\right)$
$v_{3}=\frac{u+a\left(t_{1}+t_{2}\right)+u+a\left(t_{1}+t_{2}+t_{3}\right)}{2}$
$=u+at_{1}+at_{2}+\frac{a}{2}t_{3}\quad\ldots\left(iii\right)$
Subtract $\left(i\right) from \left(ii\right)$, we get
$v_{2}-v_{1}=\frac{a}{2}\left(t_{1}+t_{2}\right)\quad\ldots\left(iv\right)$
Subtract $\left(ii\right)$ from $\left(iii\right)$, we get
$v_{3}-v_{2}=\frac{a}{2}\left(t_{2}+t_{3}\right)\quad\ldots\left(v\right)$
Divide $\left(iv\right)$ by $\left(v\right)$, we get
$\frac{v_{2}-v_{1}}{v_{3}-v_{2}}=\frac{\left(t_{1}+t_{2}\right)}{\left(t_{2}+t_{3}\right)}$ or $\frac{v_{1}-v_{2}}{v_{2}-v_{3}}=\frac{\left(t_{1}+t_{2}\right)}{\left(t_{2}+t_{3}\right)}$